PHP mysql相册允许一张照片包含在多个相册中（数据库结构）

Question

I have two tables from a mysql database with the following schema:

albums: title, date_created, date_modified, albumID (primary key) photos: caption, image_url, date_taken, imageID (primary key)

I am creating a php/mysql powered online photo album, and I would like to allow a photo to be on multiple albums, so I didn't put albumID in photos as a foreign key. The only way I can do this is to create a third table that relates the imageID to albumID, but so far I have failed to do this. What should I put on my third table that relates the imageID to albumID, and will allow users to put one photo in multiple albums.

Also, how will I query it in my php code to the mysql database to display all photos from a particular album? (i am using mysqli functions) Any help will be very much appreciated! thanks in advance

Code so far:

<form enctype="multipart/form-data" method="post">
        <?php
        require_once 'config.php'; 
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    if(isset($_POST['upload'])){
        $caption = $_POST['caption'];
        $albumID = $_POST['album'];
        $file = $_FILES ['file']['name'];
        $file_type = $_FILES ['file']['type'];
        $file_size = $_FILES ['file']['size'];
        $file_tmp = $_FILES ['file']['tmp_name'];
        $random_name = rand();

        if(empty($file)){
            echo "Please enter a file <br>";
        } else{
             move_uploaded_file($file_tmp, 'uploads/'.$random_name.'.jpg');
    $ret = mysqli_prepare($mysqli, "INSERT INTO photos (caption, image_url, date_taken, imageID)
    VALUES(?, ?, NOW()), null");
    $filename = "uploads/" . $random_name . ".jpg";
    mysqli_stmt_bind_param($ret, "ss", $caption, $filename);
    mysqli_stmt_execute($ret);
    echo "Photo successfully uploaded!<br>";
    $id = mysqli_prepare($mysqli, "INSERT into album_photo (albumID, imageID) values (?, $id)");
    }
    }
    ?>
    Caption: <br>
    <input type="text" name="caption">
    <br><br>
    Select Album: <br>
    <select name="album">
    <?php
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    $result = $mysqli->query("SELECT * FROM albums");
    while ($row = $result->fetch_assoc()) {
        $albumID = $row['albumID'];
        $title = $row['title'];
        echo "<option value='$albumID'>$title</option>";
    }
    ?>
    </select>
    <br><br>
    Select Photo: <br>
    <input type="file" name="file">
    <br><br>
    <input type="submit" name="upload" value="Upload">
</form>

Answer 1

$id = $mysqli->insert_idIf you want to have a photos exist in many albums and albums to have any photos, you would need to create a many-to-many structure. Typically you would need to create a link table, with dual primary keys.

Basically you just need a table called something like album_photo with two keys. Both are primary keys in that album, which link to photo and album as foreign keys. It could look something like this:

albumID (PK)    imageID (PK)
-------         -------
 1                1
 1                2
 2                1
 2                2

By assigning these are both primary keys, this would behave a little different than singular primary keys. Rather than having a single id appear only once, they are treated as a primary pair . which means that a grouping can only appear once. Also this would make sure that a combo could exist only once.

You would query it using SQL same as any other query like so (if you want to see a specific album):

$query = "select a.albumID, p.imageID
          from album a, photos p, album_image ai
          where a.album_ID = ai.albumID
          and p.imageID = ai.imageID
          and a.album_id = ? ";

and then include your variables in your statement to replace the ?

For your insert, assuming the user is working from an existing album, the code could look something like this:

 $insert_photo= "insert into photos (caption, image_url, date_taken, imageID)
                 values (?, ?, ?, NULL)";

 $id = [connection variable]->insert_id

$insert_link = "insert into album_photo
                albumID, imageID values (?, ?)
                WHERE albumID = $albumID and imageID = $id";

where the the last ? is your current album. Note, I only included $id in there for ease of explanation, you should also probably prepare that as well, but I won't get into that here.

EDIT: Here's an example from your code (this should get you a little closer, but you are missing a few pieces, and make sure you get the albumID from the one the user chose first, and assuming you are getting this variable through a mysqli_fetch or mysqli_fetch_array into a $row ):

$albumID = $row['albumID'];
$ret = mysqli_prepare($mysqli, "INSERT INTO photos (caption, image_url, date_taken, imageID)
VALUES(?, ?, NOW(), null");
$filename = "uploads/" . $random_name . ".jpg";
$ret->bind_param("ss", $caption, $filename);
$ret->execute();

$id = $mysqli->insert_id;
$alph = $mysqli->prepare("INSERT into album_photo (albumID, imageID) values (?, ?)  ");

//here you will need to do the rest of the binding parameters, execution, etc, just as in the first query, but off of $alph instead of $ref
$alph->bind_param("ii", $albumID, $id);
$alph->execute();
$alph->close();
$ret->close();
echo "Photo successfully uploaded!<br>";

Answer 2

Have you tried using an ORM like Doctrine or Eloquent? You can use those to implement a many-to-many relationship (which is the technical name for what you're trying to do here).

This answer is a good summary for how to implement a simple many-to-many relationship in SQL.